Pairwise commuting nilpotent matrices: alternative solution needed I have a problem:


*

*Let $A_1,A_2,...,A_n$ be $n\times n$ nilpotent matrices which are commute in each pair ($A_iA_j=A_jA_i$). Prove that:


$$A_1A_2...A_n=0$$
I have got a solution by proving that $Im(A_n)$ is an invariant supspace under $A_1...A_{n-1}$, therefore we can use the induction method by considering the $n-1$ restrictions $A_1|_{Im(A_n)}$, $A_2|_{Im(A_n)}$, ... $A_{n-1}|_{Im(A_n)}$.
However I really want to find a direct proof (maybe without using the restriction of linear transformations on an invariant supspace) since I think it would be a more intuitive way to see the problem (compared to that of induction method).
 A: Here's a slight variation:

Let $A,B \in \text{Hom}(V,V)$ be commuting matrices, with $A$ nilpotent and $B$ non-zero.  Then $\text{rank}(AB) < \text{rank}(B)$.

Proof: $A(B(V)) = B(A(V)) \subset B(V)$, so that $A\mid_{Im(B)}$ takes $Im(B)$ to $Im(B)$.  If $A(Im(B)) = Im(B)$, then $A\mid_{Im(B)}$ could not be nilpotent.  
So, we conclude that $Im(AB) = A(Im(B)) \subsetneq Im(B)$.  That is, $\text{rank}(AB) < \text{rank}(B)$.
$$\square$$
Now (assuming $A_2 \cdots A_n \neq 0$), we have
$$
\operatorname{rank}(A_1 \cdots A_n) < 
\operatorname{rank}(A_2 \cdots A_n) < \cdots <
\operatorname{rank}(A_n) < 
\operatorname{rank}(I) = n
$$
The conclusion follows.

It is debatable whether this proof is essentially different from what you've outlined, but at the very least it seems clearer.  At some point, you will have to use some fact about invariant subspaces, whether or not you do so directly.
Two alternate methods that might work are using eigenvectors (assuming $\Bbb C$ is the base field) or "simultaneous upper triangularization".
A: The interesting thing about a nilpotent operator $A$ is that it is nilpotent on any invariant subspace. So $A$ cannot be invertible on any invariant subspace $\mathcal{M}$, which means that the inequality $\dim(A\mathcal{M}) < \dim(\mathcal{M})$ must be strict except for the trivial case where $\mathcal{M}=\{0\}$.
For your case, let $X$ be the whole space with dimension $n$. Then $A_{1}X$ has dimension at most $n-1$, and is invariant under $A_{2}$. $A_{2}$ is nilpotent on $A_{1}X$; so either $A_{1}X=0$ or $A_{2}A_{1}X$ has dimension at most $n-2$, etc..
